Answer:
461.88 N
Explanation:
= Weight of the swing = 800 N
= Tension force in the rope
= Horizontal force being applied by the partner
Using equilibrium of force in vertical direction using the force diagram, we get
![T Cos30 = F_{g}\\T Cos30 = 800\\T = \frac{800}{Cos30} \\\\T = 923.76 N](https://tex.z-dn.net/?f=T%20Cos30%20%3D%20F_%7Bg%7D%5C%5CT%20Cos30%20%3D%20800%5C%5CT%20%3D%20%5Cfrac%7B800%7D%7BCos30%7D%20%5C%5C%5C%5CT%20%3D%20923.76%20N)
Using equilibrium of force in horizontal direction using the force diagram, we get
![F = T Sin30\\F = (923.76) (0.5)\\F = 461.88 N](https://tex.z-dn.net/?f=F%20%3D%20T%20Sin30%5C%5CF%20%3D%20%28923.76%29%20%280.5%29%5C%5CF%20%3D%20461.88%20N)
T = 3.5 secs
Velocity (v) = g * t = 10 m/s^2 * 3.5 sec = 35 m/s
Answer:
Yes it does.
Explanation:
"The North Magnetic Pole moves over time due to magnetic changes in Earth's core.
" - Wikipedia.
It does move around as the magnetic north does.
What the question for this assessment
Answer:
the shooting angle ia 18.4º
Explanation:
For resolution of this exercise we use projectile launch expressions, let's see the scope
R = Vo² sin (2θ) / g
sin 2θ = g R / Vo²
sin 2θ = 9.8 75/35²
2θ = sin⁻¹ (0.6)
θ = 18.4º
To know how for the arrow the tree branch we calculate the height of the arrow at this point
X2 = 75/2 = 37.5 m
We calculate the time to reach this point since the speed is constant on the X axis
X = Vox t
t2 = X2 / Vox = X2 / (Vo cosθ)
t2 = 37.5 / (35 cos 18.4)
t2 = 1.13 s
With this time we calculate the height at this point
Y = Voy t - ½ g t²
Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²
Y = 6.23 m
With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch